Krdytkyu

$$sum_{n=1}^{infty}frac1{n^2}=sum_{n=1}^{infty}frac3{n^2binom{2n}n}tag1$$

Note that $(1)$ is true since the LHS equals $zeta(2)$ whereas the RHS equals $6arcsin^2 1$ which both turn out to equal $pi^2/6$ as is well-known. However I am little bit curious about how to show $(1)$ without actually evaluating both series.

I am aware of an elegant approach contributed by Markus Scheuer as an answer to Different methods to compute Basel problem. Although this answers my question partially I am looking for different attemps. For example within Jack D'Aurizio's notes there is a way proposed exploiting creative telescoping $($see page $5$f.$)$ which I am sadly speaking not able to understand completely yet.
Hence I have come across a proof of a similiar equality concerning $zeta(3)$ on AoPS given by pprime I am
confident that there are in fact other possible methods.

I would like to see attempts of proving $(1)$ beside the mentioned which do not rely on actually showing that they both equal $pi^2/6$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similiar by pprime.

EDIT I

I have found another interesting approach, again by Jack D'Aurizio, which can be found here utilizing harmonic sums and creative telescoping in combination.

EDIT II

As pointed out by Zacky on page $31$ of Jack's notes another methode can be found which makes it three possibilities provided by Jack alone. Quite impressive!

Here is a way, although I am not sure that this is what you seek to find.

Since we have
$$
frac{1}{binom{2n}{n}}
=frac{n!n!}{(2n)!}
$$
by the binomial identity, we obtain
$$
frac{1}{n^2binom{2n}{n}}
=frac{(n-1)!(n-1)!}{(2n)!}
=frac{color{purple}{Gamma(n)Gamma(n)}}{2ncolor{purple}{Gamma(n+n)}}=frac{1}{2n}color{purple}{B(n,n)}
$$
Therefore we get
begin{align*}
sum_{n=1}^infty frac{1}{n^2binom{2n}{n}}
& =frac12sum_{n=1}^infty frac{1}{n}color{purple}{B(n,n)}
=frac12sum_{n=1}^infty frac{1}{n}color{purple}{int_0^1 (x(1-x))^{n-1}dx} \
& = -frac12int_0^1 frac{1}{x(1-x)} color{blue}{left(-sum_{n=1}^infty frac{(x(1-x))^{n}}{n}right)}dx
= - frac12 int_0^1 frac{color{blue}{ln(1-x(1-x))}}{x(1-x)}dx \
& = -frac12 bigg(int_0^1 frac{ln(1-x(1-x))}{x}dx+underbrace{int_0^1 frac{ln(1-x(1-x))}{1-x}dx}_{1-x rightarrow x}bigg) \
& = - frac12left(int_0^1 frac{ln(1-x(1-x))}{x}dx +int_0^1 frac{ln(1-(1-x)x)}{x}dxright) \
& = - int_0^1 frac{ln(1-x+x^2)}{x}dx
= - int_0^1 frac{lnleft(frac{1+x^3}{1+x}right)}{x}dx \
& = int_0^1 frac{ln(1+x)}{x}dx-underbrace{int_0^1 frac{ln(1+x^3)}{x}dx}_{x^3 rightarrow x} \
& = int_0^1 frac{ln(1+x)}{x}dx -frac13 int_0^1 frac{ln(1+x)}{x}dx
=frac23int_0^1 frac{ln(1+x)}{x}dx
end{align*}
$quad quad quad quad quad quad displaystyle{
=frac13 int_0^1 frac{ln x}{x-1}dx}$$displaystyle{=-frac13sum_{n=0}^infty int_0^1 x^n ln xdx= frac13 sum_{n=0}^infty frac{1}{(n+1)^2}=frac13 sum_{n=1}^infty frac{1}{n^2}}$

As an alternative just take $x=1$ in the following relation shown by Felix Marin:
$$
sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=-int_{0}^{1} frac{ln(1-(1-t)tx)}{t} dt.
$$